Question

At moderate pressure, the compressibility factor for a particular gas is given by: $Z = 1 + 0.3p - \dfrac{{160P}}{T}\left( {p\,in\,bar\,and\,T\,in\,\,Kelvin} \right)$. what is the Boyle's temperature of this gas?
A. $533\,K$
B. $340\,K$
C. \[470K\;\]
D. \[680\,K\]

Answer 1

Hint: The Boyle temperature is the temperature at which the gas obeys Boyle’s law i.e. the value of product of pressure and volume remains constant for an appreciable range of pressure from zero pressure. Mathematically, the Boyle temperature may be defined as the temperature at which $\mathop {\lim }\limits_{P \to 0} {\left[ {\dfrac{{\delta \left( {PV} \right)}}{{\delta T}}} \right]_T} = 0$

Complete step by step answer:
Compressibility factor shows the deviations from ideal behaviour. It is represented by letter Z which is defined as-
$Z = \dfrac{{PV}}{{{{\left( {PV} \right)}_{ideal}}}}$
For an ideal gas $340\,K$ under all conditions of temperature and pressure .The deviation of Z from unity is, thus a measure of the imperfection of the gas under consideration. At extremely low pressures, all the gases are known to have Z close to unity which means that the gases behave almost ideally. At very high pressures, all the gases have Z more than unity indicating that the gases are less comprisable than the ideal gas. This is due to the fact that at high pressures, the molecular repulsive forces are dominant. The compressibility factor goes on decreasing with increase in pressure passes through a minimum point and then begins to increase in pressure.

Value of $Z = 1$ at Boyle’s temperatures
$\begin{array}{l}
Z = 1 + 0.3p - \dfrac{{160P}}{T}\\
 = 0.3p - \dfrac{{160p}}{T} = 0\,\,\,\,\,\,(Z = 1)\\
\therefore 0.3 = \dfrac{{160}}{T}\\
T = \dfrac{{160}}{{0.3}} = 533.33K
\end{array}$

The correct option is A.

Note:
Compressibility factor shows the deviations from ideal behaviour. It is represented by letter Z which is defined as-
$Z = \dfrac{{PV}}{{{{\left( {PV} \right)}_{ideal}}}} = \dfrac{{PV}}{{nRT}} = \dfrac{{P{V_m}}}{{RT}}$where ${V_M} = \dfrac{V}{n}$ ${V_M} = \dfrac{V}{n}$is the molar volume i.e. the volume occupied by one mole of gas.